Some Basic Theorems on Probability

The classical probability problems solved so far mostly involved mutually exclusive events, using the simple additivity rule $P(A\text{ or }B)=P(A\cup B)=P(A)+P(B)$. But when two events are mutually INCLUSIVE (they can occur together), that simple additivity rule over-counts the outcomes that lie in both -- a separate formula, the Addition Theorem, is needed. This section derives that theorem and three supporting results, all proved directly from the three axioms of probability.

Theorem 12.3 -- probability of the impossible event. $P(\varnothing)=0$. Proof: the impossible event $\varnothing$ contains no sample point, so $S=S\cup\varnothing$ with $S,\varnothing$ mutually exclusive; by axiom $[P_2]$, $P(S)=P(S)+P(\varnothing)$, and since $P(S)=1\ne0$ this forces $P(\varnothing)=0$.

Theorem 12.4 -- the complement rule. If $\bar A$ is the complementary event of $A$, then $P(\bar A)=1-P(A)$. Proof: $A\cup\bar A=S$, and $A,\bar A$ are mutually exclusive, so by $[P_2]$: $P(A)+P(\bar A)=P(S)=1$, giving $P(\bar A)=1-P(A)$ (equivalently $P(A)=1-P(\bar A)$).

Theorem 12.5 -- the only-$A$ rule. For any two events $A,B$: $P(A\cap\bar B)=P(A)-P(A\cap B)$. Proof: from the Venn diagram, $(A\cap\bar B)\cup(A\cap B)=A$, and the two pieces on the left are mutually exclusive, so $P(A\cap\bar B)+P(A\cap B)=P(A)$, giving the result.

Theorem 12.6 -- the Addition Theorem on Probability. For any two events $A,B$:

Proof: $A\cup B=(A\cap\bar B)\cup B$ (the part of $A$ outside $B$, together with all of $B$), and these two pieces are mutually exclusive, so $P(A\cup B)=P(A\cap\bar B)+P(B)$; substituting Theorem 12.5's expression for $P(A\cap\bar B)$ gives $P(A\cup B)=[P(A)-P(A\cap B)]+P(B)=P(A)+P(B)-P(A\cap B)$.

Note 12.4. (i) The Addition Theorem extends to three events: $P(A\cup B\cup C)=[P(A)+P(B)+P(C)]-[P(A\cap B)+P(B\cap C)+P(C\cap A)]+P(A\cap B\cap C)$. (ii) Its complement gives the probability that NONE of $A,B,C$ occurs: $P(\bar A\cap\bar B\cap\bar C)=1-P(A\cup B\cup C)$. …